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You will need to specify a lot more information about the problem. Are your sensors reporting the heat at steady state? At some finite time after heat started flowing from the point? Is that time known? What is a heat source? Is it a single infinitesimal point? Several discrete points? A curve/area? Is it a source in that it is keeping the temperature ...


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This equation, coupled with the additional condition that $\nabla \cdot u = 0$ and some boundary conditions (otherwise the equation in underdetermined) is exactly the Navier-Stokes equation for incompressible fluid dynamics. Due to its importance in theoretical and computational study of fluids, existence and smoothness of its solutions is extremely ...


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The setup For completeness I want to state the parts from the book that are needed: We are given a (in general nonlinear) operator $$A: W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^*,$$ which is coercive (see Lemma 2.35 in the given literature). At that, coercivity means the existence of a mapping $\xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $\xi$ ...


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I'll introduce a bit of notation to make the explanation smoother. Let's start with a coordinate system $u^i=(u^1,u^2,u^3)$. The vector field with respect to this notation can be written $\mathbf x = \mathbf x (u^1,u^2,u^3) = \mathbf x (u^i)$. I'll assume that we work in Euclidean space, it's just the coordinate system that is curvilinear. So we can express ...


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Define $u_k=x-k\lambda$. Then $$ \begin{align*} \int_{0}^\lambda \sum_{k=-\infty}^\infty e^{-\left( \frac{x-k\lambda}{\sigma}\right)^2}dx &= \sum_{k=-\infty}^\infty \int_{0}^\lambda e^{-\left( \frac{x-k\lambda}{\sigma}\right)^2}dx\\ &= \sum_{k=-\infty}^\infty \int_{-k\lambda}^{-(k-1)\lambda} e^{-\left( \frac{u_k}{\sigma}\right)^2}du_k\\ &= ...


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Start changing the orders $$\int_{x = 0}^{x = \lambda} \sum_{k=-\infty}^\infty e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx=\sum_{k=-\infty}^\infty\int_{x = 0}^{x = \lambda} e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx$$ and now use, from Gaussian distribution,$$\int e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx=\frac{1}{2} \sqrt{\pi } ...


1

The characteristic curves would be $3x^2+2y^2 = C$. Hence, $u(x,y) = f(3x^2+2y^2)$ and $u(x,x) = f(5x^2) = e^{x^2}$. Hence, $f(x) = e^{x/5}$ and $u(x,y) = e^{\frac{3x^2+2y^2}{5}}$


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Let $\begin{cases}p=x^2\\q=y^2\end{cases}$ , Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=2x\dfrac{\partial u}{\partial p}$ $\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial ...


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Your numerical methods will generally converge to a weak solution, which will also be a strong solution if one exists. Here the weak solution is the zero function, and there is no strong solution (which should be unsurprising, since the forcing on the right side is not continuous). You can check that the weak solution is the zero function quite easily, ...


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You can do this. Let $v_\lambda(x)=u(\lambda x)\,,$ then $v_\lambda$ is also harmonic if $u$ is, so: $\displaystyle \Delta v_\lambda=0\implies\int_{B_r(x)}\Delta v_\lambda \,dV=0\implies \int_{\partial B_r(x)}\frac{\partial v_\lambda}{\partial r} \,r^{n-1}\,dS=0\implies \int_{\partial B_r(x)}\frac{\partial v_\lambda}{\partial \lambda} \,r^{n-1}\,dS=0\implies ...


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Notice $u_t\cong 0$ on $\partial \Omega$!


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You are correct that from the two boundary conditions, we get $v(x)=b$. But how to determine the value of that constant $b$? If you solve the initial-boundary value problem for $u(x,t)$ (by separation of variables) and take the limit as $t\to\infty$, you will see that every term in the solution vanishes except the leading (constant) term. So you have ...


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$y^{\prime}$ is an argument of $L$, so it makes perfect sense to (partial) differentiate with respect to it. Go through your question and mentally replace every instance of $y^{\prime}$ with $z$ and see if it makes more sense.


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These bump functions, or test functions, are extremely important in distribution theory. They can be constructed using partitions of unity. I didn't find any good references online in this context, but if you can get hold of Hörmanders ``The Analysis of Linear Partial Differential Operators I'', then it's an excellent reference.


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Let $B=B(0,1/2)$ and we recall the definition of weak partial derivative. We say a function $g\in L^1_{\text{loc}}(B)$ is weak derivative of $f$ if for any $\phi\in C_c^\infty (B)$ we have $$\int_B f\partial_i \phi\,dx=-\int_B g\,\phi\,dx $$ So we need to find out what is $g$ in your question. Let us suppose $x\neq 0$ and we compute the classical ...



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